T
E
X source
DERIVE source
The length of a Be\'zier curve is an elliptic integral. This is a Be\'zier curve.
x
=
a
3
t
3
+
a
2
t
2
+
a
1
t
+
x
0
y
=
b
3
t
3
+
b
2
t
2
+
b
1
t
+
y
0
The control points of a Be\'zier curve are:
(
x
0
,
y
0
) (
x
1
,
y
1
) (
x
2
,
y
2
) (
x
3
,
y
3
)
This is the curve for those control points.
a
1
=3 (
x
1
-
x
0
)
a
2
=3 (
x
0
-
2
x
1
+
x
2
)
a
3
=
-
x
0
+3
x
1
-
3
x
2
+
x
3
b
1
=3 (
y
1
-
y
0
)
b
2
=3 (
y
0
-
2
y
1
+
y
2
)
b
3
=
-
y
0
+3
y
1
-
3
y
2
+
y
3
The length of the Be\'zier curve from 0 to
u
is:
ó
õ
u
0
Ö
(
a
1
+2
a
2
t
+3
a
3
t
2
)
2
+(
b
1
+2
b
2
t
+3
b
3
t
2
)
2
dt
The control points of an example Be\'zier curve are:
(
x
0
=6,
y
0
=8) (
x
1
=1,
y
1
=10) (
x
2
=7,
y
2
=3) (
x
3
=4,
y
3
=4)
Solve for
a
1
,
a
2
,
a
3
,
b
1
,
b
2
,
b
3
.
a
1
=
-
15
a
2
=33
a
3
=
-
20
b
1
=6
b
2
=
-
27
b
3
=17
The Be\'zier curve for this example is:
x
=
-
20
t
3
+33
t
2
-
15
t
+6
y
=17
t
3
-
27
t
2
+6
t
+8
For this example the length is:
3
ó
õ
1
0
Ö
689
t
4
-
1492
t
3
+1076
t
2
-
292
t
+29
dt
Numerical integration gives 7.237223368328592885826619210956022413967067986017504163362886516
DERIVE factored the polynomial:
689
t
4
-
1492
t
3
+1076
t
2
-
292
t
+29=
689
æ
ç
ç
ç
è
t
+
æ
ú
Ö
____
Ö
6005
1378
+
23377
474721
-
373
689
+
i
æ
ç
ç
è
æ
Ö
[6005]
1378
-
23377
474721
+
7
689
ö
÷
÷
ø
ö
÷
÷
÷
ø
æ
ç
ç
ç
è
t
+
æ
ú
Ö
____
Ö
6005
1378
+
23377
474721
-
373
689
-
i
æ
ç
ç
è
æ
Ö
[6005]
1378
-
23377
474721
+
7
689
ö
÷
÷
ø
ö
÷
÷
÷
ø
æ
ç
ç
ç
è
t
-
æ
ú
Ö
____
Ö
6005
1378
+
23377
474721
-
373
689
+
i
æ
ç
ç
è
æ
Ö
[6005]
1378
-
23377
474721
-
7
689
ö
÷
÷
ø
ö
÷
÷
÷
ø
æ
ç
ç
ç
è
t
-
æ
ú
Ö
____
Ö
6005
1378
+
23377
474721
-
373
689
+
i
æ
ç
ç
è
7
689
-
æ
Ö
[6005]
1378
-
23377
474721
ö
÷
÷
ø
ö
÷
÷
÷
ø
Now define values
k
,
m
,
a
,
b
:
k
=3
___
Ö
689
m
=[1,1,1,1]
b
=[
-
1,
-
1,
-
1,
-
1]
e
=[
e
1
,
e
2
,
e
3
,
e
4
]
a
=[0.216589
-
0.0937743
i
,0.216589+0.0937743
i
, 0.866139
-
0.0734550
i
,0.866139+0.0734550
i
]
Do a partial fraction expansion.
I
(
m
)=
D
219(1,
m
,4,
a
,
b
,
e
)
I
(
m
)=
I
(4
e
1
)+(1.29909+0.375097
i
)
I
(3
e
1
)+(0.383342+0.365466
i
)
I
(2
e
1
)
-
(0.0228475
-
0.0784923
i
)
I
(
e
1
)
Reduce
I
(4
e
1
).
R
35(0,1,4,
a
,
b
,
e
)=0
A
(2
e
1
+
e
2
+
e
3
+
e
4
)+3
I
(4
e
1
)+(3.24774+0.937743
i
)
I
(3
e
1
)+(0.766684+0.730933
i
)
I
(2
e
1
)
-
(0.0342712
-
0.117738
i
)
I
(
e
1
)=0
A
(2
e
1
+
e
2
+
e
3
+
e
4
)=
-
0.138815+0.00794130
i
Substitute
I
(4
e
1
).
I
(
m
)=(0.216516+0.0625162
i
)
I
(3
e
1
)+(0.127780+0.121822
i
)
I
(2
e
1
)
-
(0.0114237
-
0.0392461
i
)
I
(
e
1
)+0.0462716
-
0.00264710
i
Reduce
I
(3
e
1
).
R
35(
-
1,1,4,
a
,
b
,
e
)=0
A
(
e
1
+
e
2
+
e
3
+
e
4
)+2
I
(3
e
1
)+(1.94864+0.562645
i
)
I
(2
e
1
)
+(0.383342+0.365466
i
)
I
(
e
1
)
-
(0.0114237
-
0.0392461
i
)
I
(0)=0
A
(
e
1
+
e
2
+
e
3
+
e
4
)=
-
0.0846852
Substitute
I
(3
e
1
).
I
(
m
)=
-
0.0655894
I
(2
e
1
)
-
(0.0415000+0.0123012
i
)
I
(
e
1
)
+(0.00246348
-
0.00389163
i
)
I
(0)+0.0554395
Reduce
I
(2
e
1
).
R
35(
-
2,1,4,
a
,
b
,
e
)=0
A
(
e
2
+
e
3
+
e
4
)+
I
(2
e
1
)+(0.0114237
-
0.0392461
i
)
I
(
-
e
1
)
+(0.649549+0.187548
i
)
I
(
e
1
)=0
A
(
e
2
+
e
3
+
e
4
)=
-
0.949300
-
0.327220
i
Substitute
I
(2
e
1
).
I
(
m
)=(0.000749279
-
0.00257413
i
)
I
(
-
e
1
)+0.00110358
I
(
e
1
)
+(0.00246348
-
0.00389163
i
)
I
(0)
-
0.00682448
-
0.0214622
i
Calculate the basic integrals numerically.
I
(
-
e
1
)=
-
17.2709+33.5142
i
I
(0)=12.0414
I
(
e
1
)=
-
3.86310
-
1.12917
i
Substitute the values of the integrals.
I
(
m
)=0.0919054
k
I
(
m
)=7.23722
That is a good match.
Jim FitzSimons Mailto:cherry@neta.com
File translated from T
E
X by
T
T
H
, version 3.86.
On 25 Nov 2009, 02:45.